A triangle $DeltaABC$ has sides a, b, and c. The angle between sides a and b is $pi/12$ and the angle between sides b and c is $pi/12$ . If side b has a length of 12, what is the area of the triangle?

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Answer 1 · 2017-06-11T06:55:54

In $DeltaABC$

$/_C=/_A=pi/12 and " side "b=12$

So $/_B=pi-2*pi/12=(5pi)/6$

Now by rule of sine

$a/sinA=b/sinB$

Now area of the

$DeltaABC=1/2axxbxxsinC$

$=1/2b/sinBxxsinAxxbxxsinC$

$=b^2/2xx(sinAsinC)/sinB$

$=b^2/2xx(sin(pi/12)sin(pi/12))/sin((5pi)/6)$

$=12^2/4xx(2sin^2(pi/12))/sin(pi-pi/6)$

$=36xx(1-cos(pi/6))/sin(pi/6)$

$=36xx(1-sqrt3/2)/(1/2)$

$=36(2-sqrt3)$ squnit

A triangle DeltaABCDeltaABC has sides a, b, and c. The angle between sides a and b is pi/12pi/12 and the angle between sides b and c is pi/12pi/12. If side b has a length of 12, what is the area of the triangle?